What letter comes next y, b, w,d,u,f,

A candidate for a US Representative seat from Indiana hires a polling firm to gauge her percentage of support among voters in he

UkoKoshka [18]

Answer:

a) The minimum sample size is 601.

b) The minimum sample size is 2401.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

The margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

For this problem, we have that:

We dont know the true proportion, so we use $\pi = 0.5$ , which is when we are are going to need the largest sample size.

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

a. If a 95% confidence interval with a margin of error of no more than 0.04 is desired, give a close estimate of the minimum sample size that will guarantee that the desired margin of error is achieved. (Remember to round up any result, if necessary.)

This is n for which M = 0.04. So

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.04 = 1.96\sqrt{\frac{0.5*0.5}{n}}$

$0.04\sqrt{n} = 1.96*0.5$

$\sqrt{n} = \frac{1.96*0.5}{0.04}$

$(\sqrt{n})^2 = (\frac{1.96*0.5}{0.04})^2$

$n = 600.25$

Rounding up

The minimum sample size is 601.

b. If a 95% confidence interval with a margin of error of no more than 0.02 is desired, give a close estimate of the minimum sample size necessary to achieve the desired margin of error.

Now we want n for which M = 0.02. So

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.02 = 1.96\sqrt{\frac{0.5*0.5}{n}}$

$0.02\sqrt{n} = 1.96*0.5$

$\sqrt{n} = \frac{1.96*0.5}{0.02}$

$(\sqrt{n})^2 = (\frac{1.96*0.5}{0.02})^2$

$n = 2401$

The minimum sample size is 2401.

4 0

3 years ago

In science class, Logan estimates the volume of a sample to be 42 mL. The actual volume of the sample is 48 mL. Find the percent

galina1969 [7]

Answer:

Percent of error of Logan's Estimate is 12.5%

Step-by-step explanation:

Given:

Actual Volume of sample =48 mL

Estimated Volume of sample = 42 mL

We need to find the percent errors of Logan estimate.

To calculate percent of error below are the steps.

Step 1: Subtract the Actual value from the estimated value.

Step 2: Take the absolute value of step 1

Step 3: Divide that answer by the Actual value.

Step 4: Multiply that answer by 100 and add the % symbol to express the answer as a percentage.

Now Substituting the values we get;

Step 1: To Subtract Actual Volume with Estimated Volume

Estimated Volume of sample - Actual Volume of sample = 42 - 48 = -6 mL

Step 2: Find Absolute value we get;

$|-6| = 6$

Step 3 : Divide that answer by the Actual volume.

$\frac{6}{48}=0.125$

Step 4: Multiply that answer by 100 and add the % symbol to express the answer as a percentage.

$0.125\times 100 = 12.5\%$

Hence Percent of error of Logan's Estimate is 12.5%

5 0

4 years ago

What is the length of the base of a right triangle with an area of 20 square meters and a height of 4 meters? 5 m 10 m 80 m 160

o-na [289]

You find the area of a triangle by using 1/2*b*h where b is the base and h is the height. since we already know the area is 20 and the height is for you plus them in so the problem would be 20= 1/2 (b) (4) then you multiply what you know together to get 20= 2b and then isolate the b by dividing both sides by two so the base should be 10 m

8 0

3 years ago

Read 2 more answers

An obtuse angle measures between:

velikii [3]

Anything over 90 and bellow 180 degrees is a obtuse.

5 0

3 years ago

Read 2 more answers

It's to hard and I can't figure it out

maria [59]

Answer: